\(\int (a+b \sin (e+f x))^m (c+d \sin (e+f x)) \, dx\) [810]

Optimal result

Integrand size = 23, antiderivative size = 230 \[ \int (a+b \sin (e+f x))^m (c+d \sin (e+f x)) \, dx=-\frac {\sqrt {2} d \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-1-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^{1+m} \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-1-m}}{b f \sqrt {1+\sin (e+f x)}}-\frac {\sqrt {2} (b c-a d) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m}}{b f \sqrt {1+\sin (e+f x)}} \] Output:

-2^(1/2)*d*AppellF1(1/2,-1-m,1/2,3/2,b*(1-sin(f*x+e))/(a+b),1/2-1/2*sin(f* 
x+e))*cos(f*x+e)*(a+b*sin(f*x+e))^(1+m)*((a+b*sin(f*x+e))/(a+b))^(-1-m)/b/ 
f/(1+sin(f*x+e))^(1/2)-2^(1/2)*(-a*d+b*c)*AppellF1(1/2,-m,1/2,3/2,b*(1-sin 
(f*x+e))/(a+b),1/2-1/2*sin(f*x+e))*cos(f*x+e)*(a+b*sin(f*x+e))^m/b/f/(1+si 
n(f*x+e))^(1/2)/(((a+b*sin(f*x+e))/(a+b))^m)
 

Mathematica [A] (warning: unable to verify)

Time = 0.47 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.87 \[ \int (a+b \sin (e+f x))^m (c+d \sin (e+f x)) \, dx=\frac {\sec (e+f x) \sqrt {-\frac {b (-1+\sin (e+f x))}{a+b}} \sqrt {\frac {b (1+\sin (e+f x))}{-a+b}} (a+b \sin (e+f x))^{1+m} \left ((b c-a d) (2+m) \operatorname {AppellF1}\left (1+m,\frac {1}{2},\frac {1}{2},2+m,\frac {a+b \sin (e+f x)}{a-b},\frac {a+b \sin (e+f x)}{a+b}\right )+d (1+m) \operatorname {AppellF1}\left (2+m,\frac {1}{2},\frac {1}{2},3+m,\frac {a+b \sin (e+f x)}{a-b},\frac {a+b \sin (e+f x)}{a+b}\right ) (a+b \sin (e+f x))\right )}{b^2 f (1+m) (2+m)} \] Input:

Integrate[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]),x]
 

Output:

(Sec[e + f*x]*Sqrt[-((b*(-1 + Sin[e + f*x]))/(a + b))]*Sqrt[(b*(1 + Sin[e 
+ f*x]))/(-a + b)]*(a + b*Sin[e + f*x])^(1 + m)*((b*c - a*d)*(2 + m)*Appel 
lF1[1 + m, 1/2, 1/2, 2 + m, (a + b*Sin[e + f*x])/(a - b), (a + b*Sin[e + f 
*x])/(a + b)] + d*(1 + m)*AppellF1[2 + m, 1/2, 1/2, 3 + m, (a + b*Sin[e + 
f*x])/(a - b), (a + b*Sin[e + f*x])/(a + b)]*(a + b*Sin[e + f*x])))/(b^2*f 
*(1 + m)*(2 + m))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3235, 3042, 3144, 156, 155}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (c+d \sin (e+f x)) (a+b \sin (e+f x))^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (c+d \sin (e+f x)) (a+b \sin (e+f x))^mdx\)

\(\Big \downarrow \) 3235

\(\displaystyle \frac {(b c-a d) \int (a+b \sin (e+f x))^mdx}{b}+\frac {d \int (a+b \sin (e+f x))^{m+1}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(b c-a d) \int (a+b \sin (e+f x))^mdx}{b}+\frac {d \int (a+b \sin (e+f x))^{m+1}dx}{b}\)

\(\Big \downarrow \) 3144

\(\displaystyle \frac {(b c-a d) \cos (e+f x) \int \frac {(a+b \sin (e+f x))^m}{\sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{b f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}+\frac {d \cos (e+f x) \int \frac {(a+b \sin (e+f x))^{m+1}}{\sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{b f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}\)

\(\Big \downarrow \) 156

\(\displaystyle \frac {(b c-a d) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \int \frac {\left (\frac {a}{a+b}+\frac {b \sin (e+f x)}{a+b}\right )^m}{\sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{b f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}+\frac {d (a+b) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \int \frac {\left (\frac {a}{a+b}+\frac {b \sin (e+f x)}{a+b}\right )^{m+1}}{\sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{b f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}\)

\(\Big \downarrow \) 155

\(\displaystyle -\frac {\sqrt {2} (b c-a d) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b f \sqrt {\sin (e+f x)+1}}-\frac {\sqrt {2} d (a+b) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m-1,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b f \sqrt {\sin (e+f x)+1}}\)

Input:

Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]),x]
 

Output:

-((Sqrt[2]*(a + b)*d*AppellF1[1/2, 1/2, -1 - m, 3/2, (1 - Sin[e + f*x])/2, 
 (b*(1 - Sin[e + f*x]))/(a + b)]*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(b*f 
*Sqrt[1 + Sin[e + f*x]]*((a + b*Sin[e + f*x])/(a + b))^m)) - (Sqrt[2]*(b*c 
 - a*d)*AppellF1[1/2, 1/2, -m, 3/2, (1 - Sin[e + f*x])/2, (b*(1 - Sin[e + 
f*x]))/(a + b)]*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(b*f*Sqrt[1 + Sin[e + 
 f*x]]*((a + b*Sin[e + f*x])/(a + b))^m)
 

Defintions of rubi rules used

rule 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ 
(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, 
 m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 156

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p 
]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p])   Int[(a + b*x)^m*(c + d*x)^n*Si 
mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
& GtQ[Simplify[b/(b*c - a*d)], 0] &&  !GtQ[Simplify[b/(b*e - a*f)], 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3144

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + 
d*x]/(d*Sqrt[1 + Sin[c + d*x]]*Sqrt[1 - Sin[c + d*x]])   Subst[Int[(a + b*x 
)^n/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Sin[c + d*x]], x] /; FreeQ[{a, b, c, 
d, n}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*n]
 

rule 3235

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)/b   Int[(a + b*Sin[e + f*x])^m, 
 x], x] + Simp[d/b   Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, 
b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
 

Maple [F]

\[\int \left (a +b \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )d x\]

Input:

int((a+b*sin(f*x+e))^m*(c+d*sin(f*x+e)),x)
 

Output:

int((a+b*sin(f*x+e))^m*(c+d*sin(f*x+e)),x)
 

Fricas [F]

\[ \int (a+b \sin (e+f x))^m (c+d \sin (e+f x)) \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )} {\left (b \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:

integrate((a+b*sin(f*x+e))^m*(c+d*sin(f*x+e)),x, algorithm="fricas")
 

Output:

integral((d*sin(f*x + e) + c)*(b*sin(f*x + e) + a)^m, x)
 

Sympy [F(-1)]

Timed out. \[ \int (a+b \sin (e+f x))^m (c+d \sin (e+f x)) \, dx=\text {Timed out} \] Input:

integrate((a+b*sin(f*x+e))**m*(c+d*sin(f*x+e)),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (a+b \sin (e+f x))^m (c+d \sin (e+f x)) \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )} {\left (b \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:

integrate((a+b*sin(f*x+e))^m*(c+d*sin(f*x+e)),x, algorithm="maxima")
 

Output:

integrate((d*sin(f*x + e) + c)*(b*sin(f*x + e) + a)^m, x)
 

Giac [F]

\[ \int (a+b \sin (e+f x))^m (c+d \sin (e+f x)) \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )} {\left (b \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:

integrate((a+b*sin(f*x+e))^m*(c+d*sin(f*x+e)),x, algorithm="giac")
 

Output:

integrate((d*sin(f*x + e) + c)*(b*sin(f*x + e) + a)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+b \sin (e+f x))^m (c+d \sin (e+f x)) \, dx=\int {\left (a+b\,\sin \left (e+f\,x\right )\right )}^m\,\left (c+d\,\sin \left (e+f\,x\right )\right ) \,d x \] Input:

int((a + b*sin(e + f*x))^m*(c + d*sin(e + f*x)),x)
 

Output:

int((a + b*sin(e + f*x))^m*(c + d*sin(e + f*x)), x)
 

Reduce [F]

\[ \int (a+b \sin (e+f x))^m (c+d \sin (e+f x)) \, dx=\left (\int \left (\sin \left (f x +e \right ) b +a \right )^{m}d x \right ) c +\left (\int \left (\sin \left (f x +e \right ) b +a \right )^{m} \sin \left (f x +e \right )d x \right ) d \] Input:

int((a+b*sin(f*x+e))^m*(c+d*sin(f*x+e)),x)
 

Output:

int((sin(e + f*x)*b + a)**m,x)*c + int((sin(e + f*x)*b + a)**m*sin(e + f*x 
),x)*d